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12 tháng 3 2020
Ta có: \(\frac{1}{5!}=\frac{1}{1\cdot2\cdot3\cdot4\cdot5}< \frac{1}{3\cdot4\cdot5}\)
\(\frac{1}{6!}< \frac{1}{1\cdot2\cdot3\cdot4\cdot5\cdot6}< \frac{1}{4\cdot5\cdot6}\)
..............
\(\frac{1}{2019!}=\frac{1}{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2019}< \frac{1}{2017\cdot2018\cdot209}\)
Do đó
\(C< 1+\frac{1}{2}+\frac{1}{2\cdot3\cdot4}+\frac{1}{4\cdot5\cdot6}+....+\frac{1}{2017\cdot2018\cdot2019}\)
\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{3-1}{1\cdot2\cdot3}+\frac{4-2}{2\cdot3\cdot4}+.....+\frac{2019-2017}{2017\cdot2018\cdot2019}\right)\)
\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2018\cdot2019}\right)< \frac{3}{2}+\frac{1}{2}\cdot\frac{1}{1\cdot2}\)
\(\Rightarrow C< \frac{7}{4}\)
Nguồn: Nock Nock
K
22 tháng 2 2020
\(C=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)
\(=\frac{1}{1}+\frac{1}{1.2}+\frac{1}{1.2.3}+...+\frac{1}{1.2.3...2019}\)
\(=\frac{1}{1}+\frac{1}{1}.\frac{1}{2}+\frac{1}{1}.\frac{1}{2}.\frac{1}{3}+...+\left(\frac{1}{1}.\frac{1}{2}.\frac{1}{3}...\frac{1}{2018}.\frac{1}{2019}\right)\)
\(=\left(1.1.1....1.1\right)+\left(\frac{1}{2}.\frac{1}{2}.\frac{1}{2}...\frac{1}{2}.\frac{1}{2}\right)+\left(\frac{1}{3}.\frac{1}{3}.\frac{1}{3}...\frac{1}{3}.\frac{1}{3}\right)+...+\left(\frac{1}{2018}.\frac{1}{2018}\right)+\frac{1}{2019}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}+\frac{1}{2019}\)
Nhận xét rằng:
\(1< \frac{7}{8076};2< \frac{7}{8076};3< \frac{7}{8076};...;\frac{1}{1154}>\frac{7}{8076};\frac{1}{1155}>\frac{7}{8076};...;\frac{1}{2018}>\frac{7}{8076};\frac{1}{2019}>\frac{7}{8076}\)
Do đó:
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2018}+\frac{1}{2019}>\frac{7}{8076}+\frac{7}{8076}+...+\frac{7}{8076}\)
Vì tổng C có 2019 số hạng, suy ra \(C>2019.\frac{7}{8076}=\frac{7}{4}\)
NC
0
GT
0
PU
1
8 tháng 11 2019
\(B=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2019}}\)
\(\Rightarrow5B=1+\frac{1}{5}+...+\frac{1}{5^{2018}}\)
\(\Rightarrow5B-B=\left(1+\frac{1}{5}+...+\frac{1}{5^{2018}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2019}}\right)\)
\(\Rightarrow4B=1-\frac{1}{5^{2019}}< 1\)
\(\Rightarrow B< \frac{1}{4}\)
N
2
QH
15 tháng 6 2020
A= (10^2019+7)/(10^2019 + 1) = 1+ (6 / 10 ^2019+1)
B = ( 10 ^ 2020 +9) / ( 10 ^2020 +3) = 1 +( 6 / 10^ 2020 +3)
A -B = (6 / 10 ^2019+1) - (6 / 10^2020 +3) >0
=> A > B
NT
Câu 1: Cho A.= \(\frac{7^{2018}+1}{7^{2019}+1}\)Và B=\(\frac{7^{2019}+1}{7^{2019}+1}\)
So sánh A và B
2
7 tháng 8 2019
\(A=\frac{7^{2018}+1}{7^{2019}+1}\)
\(\Rightarrow7A=\frac{7^{2019}+7}{7^{2019}+1}=1+\frac{6}{7^{2019}+1}\)
\(B=\frac{7^{2019}+1}{7^{2020}+1}\)
\(\Rightarrow7B=\frac{7^{2020}+7}{7^{2020}+1}\)
\(\Rightarrow7B=1+\frac{6}{7^{2020}+1}\)
Vì 7 ^ 2019 < 7 ^ 2020 => 7 ^ 2019 + 1 < 7 ^ 2020 + 1
=> 6 / ( 7 ^ 2019 + 1 ) > 6 / ( 7 ^ 2020 + 1 )
=> 1 + 6 / ( 7 ^ 2019 + 1 ) > 1 + 6 / ( 7 ^ 2020 + 1 )
=> 7A > 7B
Vì A , B > 0
Nên A > B
BY
7 tháng 8 2019
Vì \(7^{2018}< 7^{2019}\)nên \(7^{2018}+1< 7^{2019}+1\)
\(\Rightarrow\frac{7^{2018}+1}{7^{2019}+1}< \frac{7^{2019}+1}{7^{2019}+1}\)
Hay A < B
Chúc bạn học tốt ! Nguyễn Thi An Na
OM
2
AH
Akai Haruma
Giáo viên
11 tháng 3 2020
Lời giải:
Ta có:
\(\frac{1}{5!}=\frac{1}{1.2.3.4.5}< \frac{1}{3.4.5}\)
\(\frac{1}{6!}< \frac{1}{4.5.6}\)
.........
\(\frac{1}{2019!}< \frac{1}{2017.2018.2019}\)
Do đó:
\(C< 1+\frac{1}{2}+\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{2017.2018.2019}\)
\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{2019-2017}{2017.2018.2019}\right)\)
\(C< \frac{3}{2}+\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2018.2019}\right)< \frac{3}{2}+\frac{1}{2}.\frac{1}{1.2}\)
\(C< \frac{7}{4}\)
BD
16 tháng 7 2023
a) Ta có:
2A=2.(12+122+123+...+122020+122021)2�=2.12+122+123+...+122 020+122 021
2A=1+12+122+123+...+122019+1220202�=1+12+122+123+...+122 019+122 020
Suy ra: 2A−A=(1+12+122+123+...+122019+122020)2�−�=1+12+122+123+...+122 019+122 020
−(12+122+123+...+122020+122021)−12+122+123+...+122 020+122 021
Do đó A=1−122021<1�=1−122021<1.
Lại có B=13+14+15+1360=20+15+12+1360=6060=1�=13+14+15+1360=20+15+12+1360=6060=1.
Vậy A < B.
! là gì thế